Dimensions of Hilbert Spaces confusion

[nomadreid]

If I understand it, Hilbert spaces can be finite (e.g., for spin of a particle), countably infinite (e.g., for a particle moving in space), or uncountably infinite (i.e., non-separable, e.g., QED). I am wondering about variations on this latter. The easiest uncountable to imagine is the cardinality of the continuum. What examples are there of other uncountable cardinalities in quantum physics (either more or, if assuming the negation of the continuum hypothesis, less)? Thanks.

 

[George Jones]

Which Hilbert space in QED is non-separable?

 

[nomadreid]

For example
"In QED it is (at least) the continuum of directions in 3-space. One needs this nonseparable space to define Lorentz transformations of charges states, as charged states moving in different directions are in different superselection sectors. Thus the dimension of the Hilbert space of the universe should be at least the cardinality of the continuum." from http://physics.stackexchange.com/qu...rse-have-to-be-infinite-dimensional-to-make-s, or http://en.wiktionary.org/wiki/quantum_electrodynamics

More cryptically
"quantum electrodynamics (uncountable)" in http://en.wiktionary.org/wiki/quantum_electrodynamics

Then, we have field theory:
http://en.wikipedia.org/wiki/Partition_function_(quantum_field_theory)

 

[Bill_K]

Which Hilbert space in QED is non-separable?

A quantum field is Fourier transformed into normal modes, a set of N simple harmonic oscillators, one for each value of k. Each mode represents a dimension in the Fock space. If N is finite, the Fock space is finite dimensional. If N is infinite but countable (k values are discrete), the Fock space is infinite dimensional and separable. If the allowed k values are continuous, N is infinite and uncountable, and the Fock space is non-separable.

 

[George Jones]

If N is infinite but countable (k values are discrete), the Fock space is infinite dimensional and separable. If the allowed k values are continuous, N is infinite and uncountable, and the Fock space is non-separable.

I guess my question is, when, specifically, does this actually happen, as opposed to, appear to happen?

As a much simpler example, consider the state space for single particle in one-dimensional non-relativistic quantum mechanics. I have seen some folks say, incorrectly, that this state space is non-separable because it has $\left\{ \left| x \right> \right\}$ is an uncountable basis.

For me, the mathematics is very subtle.

 

[Bill_K]

Yikes, I did get it wrong,. Let me erase everything I said and start over.

Consider first a single particle, with just one degree of freedom and one canonical pair of operators p, q or raising/lowering operators a*, a. Starting with the vacuum state |0> application of the raising operator produces a countably infinite sequence of basis states (these are all linearly independent, and a general state in the space is a linear combination of all of them) and the Hilbert space is infinite dimensional and separable.

Next, for a system with a finite number of degrees of freedom, N. Each of the N raising operators generates a sequence of states. A basis vector in the Hilbert space can be represented by specifying N occupation numbers, e.g. |n₁, n₂,... n_N> This is an infinite basis but still a countably infinite basis, and the Hilbert space is still separable.

However if N → ∞, a system with an infinite number of degrees of freedom, aka a field, the basis becomes uncountably infinite, |n₁, n₂,... > (aleph-0 to the aleph-0 power is aleph-1) and the Hilbert space is non-separable.

 

[dextercioby]

As George says, there's a confusion here which stems from incorrectly using the terminology.

The Fock space of a free quantum field is built upon countable products and sums of rigged Hilbert spaces (more precisely, distributional spaces from the 3 spaces which make up a rigged Hilbert space of a single particle).

So I would like to see a Hilbert space used in physics which is uncountable.

 

[micromass]

Yikes, I did get it wrong,. Let me erase everything I said and start over.

Consider first a single particle, with just one degree of freedom and one canonical pair of operators p, q or raising/lowering operators a*, a. Starting with the vacuum state |0> application of the raising operator produces a countably infinite sequence of basis states (these are all linearly independent, and a general state in the space is a linear combination of all of them) and the Hilbert space is infinite dimensional and separable.

Next, for a system with a finite number of degrees of freedom, N. Each of the N raising operators generates a sequence of states. A basis vector in the Hilbert space can be represented by specifying N occupation numbers, e.g. |n₁, n₂,... n_N> This is an infinite basis but still a countably infinite basis, and the Hilbert space is still separable.

However if N → ∞, a system with an infinite number of degrees of freedom, aka a field, the basis becomes uncountably infinite, |n₁, n₂,... > (aleph-0 to the aleph-0 power is aleph-1) and the Hilbert space is non-separable.

But $\aleph_0^{\aleph_0}$ is not $\aleph_1$. In fact, it is unknown which $\aleph_\alpha$ exactly it is. Under the continuum hypothesis, it is $\aleph_1$, but this is not universally accepted.

Also, you described something like a tensor product of Hilbert spaces with infinitely many factors. How exactly is this defined?
I'm not very familiar with Dirac notation, but I think that for finitely many factors, we have something like
$$<n_1n_2|m_1m_2> = <n_1|m_1><n_2|m_2>$$

But how would this be with infinitely many factors?

 

[Bill_K]

But $\aleph_0^{\aleph_0}$ is not $\aleph_1$. In fact, it is unknown which $\aleph_\alpha$ exactly it is. Under the continuum hypothesis, it is $\aleph_1$, but this is not universally accepted.

Granted, I should have just said that in any case, $\aleph_0^{\aleph_0}$ is clearly greater than $\aleph_0$, and therefore uncountable.

Also, you described something like a tensor product of Hilbert spaces with infinitely many factors. How exactly is this defined?

Not so hard. Write the basis vectors using occupation numbers, |n₁, n₂,... >.
Then <m₁, m₂ ...|n₁, n₂ ...> = 1 if and only if m₁ = n₁ and so on.

 

[rubi]

The Hilbert space of standard QFT is constructed this way: Fix a separable one-particle Hilbert space $\mathcal H$, define $\mathcal H^{\otimes 0} = \mathbb C$ and $\mathcal H^{\otimes n} := P \bigotimes_{k=1}^n \mathcal H$, where $P$ is a projector onto the symmetric or totally antisymmetric subspace. The Hilbert space of QFT is then $\mathcal F = \bigoplus_{n=0}^\infty \mathcal H^{\otimes n}$ (the definition of the inner product should be clear now). It's also clear that $\mathcal F$ is separable since $\mathcal H^{\otimes n}$ is separable and the direct sum of countably many separable spaces is separable as well. The difference between field theory and many-particle QM is that the observables become operator-valued distributions.

 

[Bill_K]

The Hilbert space of standard QFT is constructed this way: Fix a separable one-particle Hilbert space $\mathcal H$, define $\mathcal H^{\otimes 0} = \mathbb C$ and $\mathcal H^{\otimes n} := P \bigotimes_{k=1}^n \mathcal H$, where $P$ is a projector onto the symmetric or totally antisymmetric subspace. The Hilbert space of QFT is then $\mathcal F = \bigoplus_{n=0}^\infty \mathcal H^{\otimes n}$ (the definition of the inner product should be clear now). It's also clear that $\mathcal F$ is separable since $\mathcal H^{\otimes n}$ is separable and the direct sum of countably many separable spaces is separable as well. The difference between field theory and many-particle QM is that the observables become operator-valued distributions.

No, it isn't the direct sum, it's the direct product.

 

[micromass]

No, it isn't the direct sum, it's the direct product.

Reed and Simon's Methods of Mathematical physics, page 53 says that it's the direct sum.

Could you give me a reference to where the direct product of Hilbert spaces is actually defined?

 

[rubi]

No, it isn't the direct sum, it's the direct product.

It's the direct sum of ((anti-)symetrized) direct products (as i wrote). The direct products are all finite. Only the direct sum is over infinitely many spaces. It's also written here: https://en.wikipedia.org/wiki/Fock_space

 

[Fredrik]

I guess my question is, when, specifically, does this actually happen, as opposed to, appear to happen?

I have no idea, but I was pretty sure that A. Neumaier had mentioned something about it, and probably included a reference. I did a quick search and didn't find any references, but I found a post where he made a comment about it. https://www.physicsforums.com/showthread.php?p=3988874#post3988874

 

[George Jones]

I have no idea, but I was pretty sure that A. Neumaier had mentioned something about it, and probably included a reference. I did a quick search and didn't find any references, but I found a post where he made a comment about it. https://www.physicsforums.com/showthread.php?p=3988874#post3988874

In post #3, nomadreid also referenced Neumaier, but did not give a text reference. QED has been around along time, so, if one exists, I would like to see an example from a text of a standard use of non-separable Hilbert spaces in QED.

As has already been mentioned, a Fock space based on a separable Hilbert space is itself separable.

 

[Bill_K]

As has already been mentioned, a Fock space based on a separable Hilbert space is itself separable.

Again, this is simply not the case. Here's what you're missing: the nonseparability only arises when there can be an infinite number of nonzero occupation numbers. Such states cannot be reached from the vacuum state in a finite number of steps. They come into play when you're dealing with coherent states, which contain an infinite number of photons.

 

[dextercioby]

Bill, I don't understand. Non-separability has to do with whether the vector space is endowed with a countable basis or not. What is the vector space and what is the basis ?

 

[rubi]

In post #3, nomadreid also referenced Neumaier, but did not give a text reference. QED has been around along time, so, if one exists, I would like to see an example from a text of a standard use of non-separable Hilbert spaces in QED.

Neumaier is both right and wrong in the SE posting. He is right about the fact that the Hilbert space of QED can be written as a direct integral (it arises if you apply Mackey's theory of unitary representations to the Poincare group). It's not true however, that this implies that it is non-separable. For example, the space $L^2(\mathbb R) = \int_{\mathbb R}^\oplus \mathbb C \,\mathrm d \lambda(x)$ can be written as a direct integral over an uncountable domain as well (where $\lambda$ is the Lebesgue measure), yet we know that it is separable.

Again, this is simply not the case. Here's what you're missing: the nonseparability only arises when there can be an infinite number of nonzero occupation numbers. Such states cannot be reached from the vacuum state in a finite number of steps. They come into play when you're dealing with coherent states, which contain an infinite number of photons.

It's just wrong what you're saying. The Fock space is a direct sum of spaces $P\mathcal H^{\otimes n}$ and each of them has a countable basis $(\psi^n_k)_k$. The collection $(\psi^n_k)_{nk}$ is then a basis for the direct sum and it is countable since the index is in $\mathbb N^2$, which is a countable set.

A coherent state is a superposition of states with a finite number of particles. It is given by $\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} \left|n\right>$, where each $\left|n\right> \in P\mathcal H^{\otimes n}$. So a coherent state doesn't contain an infinite number of photons, but instead it's an infinite superposition of states that contain finitely many photons.

 

[Bill_K]

Bill, I don't understand. Non-separability has to do with whether the vector space is endowed with a countable basis or not. What is the vector space and what is the basis ?

Because the basis states with a finite number of nonzero occupation numbers span a subspace which IS separable. It has a countable basis. Like the rationals as a subspace of the reals.

If you look on the web for other mentions of nonseparable Hilbert spaces in QM, you'll find them in the context of coherent photon states, as I pointed out. For example this one:

...with the nonseparable Hilbert space, defined in Paper I, which is spanned by the soft-photon coherent states.

 

[rubi]

If you look on the web for other mentions of nonseparable Hilbert spaces in QM, you'll find them in the context of coherent photon states, as I pointed out. For example this one:

I agree that infinite tensor-product spaces are non-separable. However, QED isn't formulated on such a space. Instead it is formulated on a Fock space, which is separable.

 

[Bill_K]

Many of the discussions this topic that I'm finding on the web, even though they say basically the same thing, aren't peer reviewed, (e.g. stackexchange threads) and so I can't claim them as an authority. However this one from functionspace.org echoes what I've been saying, and perhaps offers further clarification, so just consider it as words I wish I'd said:

Even in quantum field theory, most of the Hilbert spaces are in fact separable, as stipulated by the Wightman axioms. However, it is sometimes argued that non-separable Hilbert spaces are also important in quantum field theory, roughly because the systems in the theory possesses an infinite number of degrees of freedom and any infinite Hilbert tensor product (of spaces of dimension greater than one) is non-separable. For instance, a bosonic field can be naturally thought of as an element of a tensor product whose factors represent harmonic oscillators at each point of space. From this perspective, the natural state space of a boson might seem to be a non-separable space. However, it is only a small separable subspace of the full tensor product that can contain physically meaningful fields (on which the observables can be defined)

.

 

[George Jones]

We seem to be going back and forth on this.

I had a vague recollection of years ago reading something relevant, so I went looking.

On page 33 of his book "Quantum Field Theory on Curved Spacetime and Black Hole Thermodynamics", Wald writes (for quantum fields in flat Minkowski spacetime)

In the previous chapter, we gave two equivalent prescriptions for constructing the Hilbert space for the quantum theory of finitely many oscillators: 1) Take the tensor product space of the Hilbert spaces for the individual oscillators. 2) Construct a Fock space based upon a Hilbert space of (complex) classical solutions. In the present case of infinitely many oscillators, this first prescription is no longer available. Although the tensor product of infinitely many Hilbert spaces can be defined (von Neumann 1938), it is "too large" to be suitable for use in a (normal type of) quantum theory in that it is not separable and the representation it provides of the canonical commutation relations is reducible.

On the other hand, the second prescription generalizes straightforwardly to the present case.

 

[strangerep]

I agree that infinite tensor-product spaces are non-separable. However, QED isn't formulated on such a space. Instead it is formulated on a Fock space, which is separable.

If I understand Kibble's construction, he was trying to introduce an improved QED formulation over the standard Fock-space formulation of QED. In the latter, one makes sense of IR divergences only at the level of cross-sections, not the S-matrix. However, in Kibble's framework he banishes IR divergences from the S-matrix to all orders of perturbation theory, albeit at the cost of having asymptotic fields which are less convenient to work with.

Certainly, Kibble's space is unitarily inequivalent to the usual QED Fock Space, but does he not settle for a particular space among an uncountable set of alternatives?

Separately, (iirc) at least one existence proof for certain low-dimensional scalar field theories relies on progressively moving to a different unitarily-inequivalent space at each order of perturbation theory and then showing that this process converges in some sense.

Sure wish DarMM would stop by...

---------------------------------
For sufficiently masochistic readers, here are references to the other Kibble papers:

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences. I. Classical Currents,
J. Math. Phys., vol 9, no. 2, (1968), p. 315.

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences.
II. Mass-Shell Singularities of Green's Functions,
Phys. Rev., vol 173, no. 5, (1968), p. 1527.

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences.
III. Asymptotic States and Reduction Formulas.,
Phys. Rev., vol 174, no. 5, (1968), p. 1882.

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences.
IV. The Scattering Operator.,
Phys. Rev., vol 175, no. 5, (1968), p. 1624.

 

[Bill_K]

I think we've pretty much got it nailed down, although I see two points that deserve further comment.

... it is only a small separable subspace of the full tensor product that can contain physically meaningful fields (on which the observables can be defined)

What's being referred to here is that for basis states that contain an infinite number of particles, the eigenvalues of the observables may well be infinite. That's why the infinite tensor product is physically realizable only on coherent states, in which the infinite number of soft photons still has finite total energy.

Although the tensor product of infinitely many Hilbert spaces can be defined (von Neumann 1938)... the representation it provides of the canonical commutation relations is reducible.

This is both important and interesting! The usual representation of the canonical commutation relations contains the vacuum state, and all other states belonging to it can be obtained from the vacuum by application of the raising/lowering operators. This is conventionally what we mean by the Fock space. But states with an infinite number of particles cannot be obtained in this way, and therefore belong to another irreducible representation. The representation over the entire Hilbert space is reducible. In fact, any two states whose occupation numbers differ at an infinite number of places belong to separate representations.

 

[DarMM]

Quantum field theories at zero-density have separable Hilbert spaces. It can be proven that the separability axiom is equivalent to the statement that in the theory all states have a finite particle number expectation value. At non-zero densities this fails and introduces a whole host of difficulties, especially with the mathematical treatment of thermal states.

Also, for non-separable Hilbert spaces, just because a group has a representation on the Hilbert space does not imply that its Lie-Algebra has a representation (failure of Stone's theorem). Physically this would mean that symmetry would not imply the existence of a conserved quantity (failure of Noether's theorem in finite density field theory)

 

[samalkhaiat]

You seem to have ignored the fact that working with non-separable/separable spaces is decided by the type of your “subsidiary condition”.

We all know that QFT contains massless gauge fields whose covariant quantization is done in indefinite metric. So, to have a sensible theory, one introduces a “subsidiary condition” by means of which the physical subspace is projected out of the indefinite-metric Hilbert space. Indeed, one can show that the physical S-matrix, defined by restricting the total S-matrix to the physical subspace, is unitary.

So let’s talk about subsidiary condition. In order to define a subspace, it must be linear with respect to states (however, non-linear subsidiary conditions were often introduced in the past because finding an appropriate one is rather difficult).

(1) Let $Q$ be certain non-trivial operator, then the totality of the states $| X \rangle$ satisfying

$$\langle X | Q | X \rangle = 0 \ , \ \ \ (1)$$

does not constitute a subspace (for example, the totality of time-like vectors is not a subspace of the Minkowski space).

The physical subspace $\mathcal{ F }$ is the set of all (physical) states $| X \rangle$ solving

$$Q | X \rangle = 0 . \ \ \ \ (2)$$

Of course, we must require that $| 0 \rangle \in \mathcal{ F }$. Therefore, we should have

$$Q | 0 \rangle = 0 \ . \ \ \ \ (3)$$

This means that $Q$ cannot be a local operator, otherwise the separating property of vacuum requires that $Q = 0$.

“Thus a subsidiary condition should be non-local one”

(2) Since $Q$ is a non-trivial operator, there must exist an operator $\phi$ such that $[ Q , \phi ] \neq 0$. Now, if $Q$ is spacetime-dependent operator, we often find

$$[ Q , \phi ] = C + R , \ \ \ (4)$$

where $C$ is a non-zero c-number and $R$ is either zero or a q-number such that
$$\exists | X \rangle \in \mathcal{ F } , \ \langle X | X \rangle \neq 0 , \ \ \langle X | R | X \rangle = 0 .$$

Notice that a Hermitian $Q$ leads to the following wrong statement

$$0 = \langle X | [ Q , \phi ] | X \rangle = C \langle X | X \rangle \neq 0 .$$

“Thus $Q$ should not be Hermitian”

However, a spacetime-dependent Hermitian $Q$ can be used (and indeed has been proposed many times so far). We can avoid the above contradiction if we admit a continuous set of states. We encounter similar situation in QM: the expectation value of $[ x , p ] = i$ in a definite eigenstate of $p$ leads to similar contradiction, but this procedure is not allowed in the case of continuous spectrum.

Thus, in order to consider a spacetime-dependent Hermitian $Q$, we must introduce a state-vector space having continuously infinite number of independent vectors, i.e. a nonseparable space. However, there are many problems with nonseparable space: It is a too large space. An infinitesimal symmetry transformation is not always represented by an infinitesimal change of a state vector, and the corresponding symmetry generator, therefore, may not exist. And, to define probability, the inner product of states must be supplemented by some integration weight with no general principle for determining it.

Therefore, we (usually) restrict the formulation of QFT to a separable space. Then, it is expected that $Q$ is either a non-local, spacetime-dependent, non-hermitian operator or a spacetime independent one (unbroken symmetry generator).

Sam

 

[tom.stoer]

Interesting.

Could you please explain where we do get a problem in QCD on a 3-torus with Weyl-gauge $A^0 = 0$ plus Gauß-law constraint $G^a(x)|\text{phys.}\rangle = 0$.

 

[DarMM]

Very good post samalkhaiat. If I remember correctly this only fails in finite density field theory, it being the only scenario where one is forced in certain cases to accept non-seperable spaces.

 

[samalkhaiat]

Interesting.

Could you please explain where we do get a problem in QCD on a 3-torus with Weyl-gauge $A^0 = 0$ plus Gauß-law constraint $G^a(x)|\text{phys.}\rangle = 0$.

The problem is this, if I want to answer your question properly, I end up writing a book. So I will only address the general problems in QCD and point you to a good review article.
The problem is how to interpret the Gauss’ law constraints,
$$G_{ a } ( x ) | X \rangle = 0 . \ \ \ (1)$$
This seems to imply that the physical states are gauge invariant, but even in QED this would lead to difficulties. The problem is that the local gauge transformations also include the global ones. In this case, the abelian charge becomes the electric charge, and physical states annihilated by this can only have zero charge. This is clearly an obstacle when we want to construct charged states. To avoid this we must have a non-trivial action of the global transformations on the physical states. To determine then what Gauss’ subsidiary condition means in terms of the gauge invariance of the system we will need to analyse how the colour charge is defined in QCD.
We would expect to identify the colour charge with the Noether charge
$$Q_{ a } = \int d^{ 3 } x \ j^{ 0 }_{ a } ( x ) = \int d^{ 3 } x \ ( J^{ 0 }_{ a } – f_{ a b c } E^{ b }_{ j } A^{ c j } ) , \ \ \ (2)$$
where $J^{ \mu }_{ a } ( x )$ is the matter field current. This charge is, by construction, time independent but it is not gauge invariant. We know that constituent quarks have a well-defined colour. So, how can we reconcile this with the gauge structure of QCD? The answer relays on taking the gauge invariance of any physical state seriously. We can show that if this charge is acting on a physical state, as defined by (1), then we have
$$Q = \frac{ 1 }{ g } \int d^{ 3 } x \ \partial^{ j } E_{ j } = \frac{ 1 }{ g } \lim_{ R \rightarrow \infty } \int_{ S^{ 2 }_{ R } } d \vec{ s } \cdot \vec{ E } .$$
Under a gauge transformation we have $E_{ j } \rightarrow U^{ -1 } E_{ j } U$ so that
$$Q \rightarrow Q^{ U } = \frac{ 1 }{ g } \lim_{ R \rightarrow \infty } \int_{ S^{ 2 }_{ R } } d \vec{ s } \cdot U^{ -1 } \vec{ E } U .$$
In order to obtain the group element from this integral, we must assume that it tends to a constant $U_{ \infty }$ in an angle-independent way. Then
$$Q_{ U } = U^{ -1 }_{ \infty } Q U_{ \infty } .$$
So, we see that the colour charge, when acting on physical states, is gauge invariant under those gauge transformations that at spatial infinity tend to a constant $U_{ \infty }$ which lies in the centre of $SU_{ c }(3)$. I will not talk about the structure of these gauge transformations, but only make the observation that when $U_{ \infty }$ is the identity element of $SU_{ c }(3)$, then the coloure charge (2) is invariant.
So, now we define the group of local gauge transformations to be those that become the identity at spatial infinity. But, we also have the global gauge transformations. Notice the important difference between the two: to define colour charge we were forced to restrict the local transformations to those which coincide with the identity at spatial infinity, this means that, in this description, the global transformations are not a special class of the local ones. Therefore, we see that in QCD we have to consider those transformations which are a combination of such local and global transformations.
We also know, due to the instanton structure, that the local gauge group is disconnected [2]. Thus, only the part of the group connected to the identity that can be generated from the infinitesimal gauge transformations and hence the Gauss’ law. So, in order for the colour charge to be well defined, the physical states, as defined by the constraint (1), are invariant under those local transformations that belong to the identity component of the group of “all” gauge transformations. Indeed, one can argue that the BRST charges give a better characterization of these states.

Okay, I think it is better to stop this unfinished business and direct you to Jackiw excellent article

[1] Jackiw, R. in Relativity, Group and Topology II, Les Houches 1983, eds., Dewitt, B. S. and Stora, R. (North- Holland, 1984)
An updated version of this is reprinted in “Current Algebra and Anomalies” World Scientific 1985.

See also the following review

[2] Lavelle, M. and McMullan, D. Phys. Rep. 279, 1(1997).

Sam

 

[samalkhaiat]

Very good post samalkhaiat. If I remember correctly this only fails in finite density field theory, it being the only scenario where one is forced in certain cases to accept non-seperable spaces.

Well, before the Gupta and Bleuler subsidiary condition appeared many people worked with non-separable space. As for finite density field theory, I believe, in those case, it is the absence of linear constraint that forces one to pick up non-separable space.

 

[tom.stoer]

samalkhaiat, thanks for the excellent summary, but I think you miss some important points.

The Gauß law is identical to the generator of gauge transformations only modulo integration by parts, i.e. surface terms; of course they vanish on the compact 3-torus.

In QCD on compact spaces it is natural that physical states are color-neutral states. And I don't see why there should be a problem at all if the theory predicts strict color-neutrality.

So there may be obstacles on R³ but not on T³.

 

[samalkhaiat]

samalkhaiat, thanks for the excellent summary, but I think you miss some important points.

Well, I missed a lot of important points. I already said I would.

The Gauß law is ... i.e. surface terms; of course they vanish on the compact 3-torus.

Do they? Can you show me how you can get rid of the “Schwinger” type terms in the commutator algebra that come from the anomalous divergence of the source current? As far as I know, with careful calculations one ends up with
$$[ i G_{ a } ( x ) , G_{ b } ( y ) ] = f_{ a b c } G_{ c } ( x ) \delta^{ 3 } ( x – y ) - i S_{ a b } ( A ; x , y ) \ \ (1)$$
It is known fact that whenever a dynamical current, i.e. a source current in a gauge theory, possesses a anomalous divergence, then also the equal-time commutators for its components must contain anomalous terms. This is because a divergence can be represented by commutator with the translation generators $P^{ \mu }$ and the commutator must be anomalous since the divergence is. Since $P^{ 0 } = H$ contains $J^{ \mu }_{ a }$, one expects that the equal-time commutator $[ J^{ \mu }_{ a } , J^{ 0 }_{ b } ]$ is anomalous. This commutator is one source of anomaly in (1). The other coming from the commutator of $( D_{ j } \delta / \delta A_{ j } )_{ a }$ with $J^{ 0 }_{ b }$. This commutator induces infinitesimal gauge transformation on any explicit dependence of $J^{ 0 }_{ b }$ on the gauge field.

In QCD on compact spaces it is natural that physical states are color-neutral states. And I don't see why there should be a problem at all if the theory predicts strict color-neutrality.

That does not help me to determine how much spin do the quarks and gluons contribute to the proton.

 

[strangerep]

See also the following review

[2] Lavelle, M. and McMullan, D. Phys. Rep. 279, 1(1997).

It's so cool that you mention these authors! Their papers are really interesting. (I've been working with some of their other papers that deal with IR issues and gauge invariance in QED.)

For the benefit of other readers, that paper is freely available here.

 

[tom.stoer]

Can you show me how you can get rid of the “Schwinger” type terms in the commutator algebra that come from the anomalous divergence of the source current? As far as I know, with careful calculations one ends up with
$$[ i G_{ a } ( x ) , G_{ b } ( y ) ] = f_{ a b c } G_{ c } ( x ) \delta^{ 3 } ( x – y ) - i S_{ a b } ( A ; x , y ) \ \ (1)$$

With gauge-invariant regularization these Schwinger terms do vanish for gauge currents. (I don't know whether this works for chiral gauge theories)

That does not help me to determine how much spin do the quarks and gluons contribute to the proton.

Of course it doesn't. I don't see why the local color gauge algebra shall be able to do that. For the color-neutral axial current things may change, but that's not a gauge current.

 

[samalkhaiat]

It's so cool that you mention these authors! Their papers are really interesting. (I've been working with some of their other papers that deal with IR issues and gauge invariance in QED.)

For the benefit of other readers, that paper is freely available here.

They understand the role played by Poincare’ generators in gauge field theories. And that is good enough for me. Plus, one of them is a good teacher.